Find the domain and range of f(x)=x²+1/ x²-5x + 6?
Answers
Answered by
0
The domain is
x
∈
(
−
∞
,
−
3
)
∪
(
−
3
,
−
2
)
∪
(
−
2
,
+
∞
)
. The range is
y
∈
(
−
∞
,
−
4
]
∪
[
0
,
+
∞
)
Explanation:
The denominator is
x
2
+
5
x
+
6
=
(
x
+
2
)
(
x
+
3
)
As the denominator must be
≠
0
Therefore,
x
≠
−
2
and
x
≠
−
3
The domain is
x
∈
(
−
∞
,
−
3
)
∪
(
−
3
,
−
2
)
∪
(
−
2
,
+
∞
)
To find the range, proceed as follows :
Let
y
=
1
x
2
+
5
x
+
6
y
(
x
2
+
5
x
+
6
)
=
1
y
x
2
+
5
y
x
+
6
y
−
1
=
0
This is a quadratic equation in
x
and the solutions are real only if the discriminant is
≥
0
Δ
=
b
2
−
4
a
c
=
(
5
y
)
2
−
4
(
y
)
(
6
y
−
1
)
≥
0
25
y
2
−
24
y
2
+
4
y
≥
0
y
2
+
4
y
≥
0
y
(
y
+
4
)
≥
0
The solutions of this inequality is obtained with a sign chart.
The range is
y
∈
(
−
∞
,
−
4
]
∪
[
0
,
+
∞
)
Answered by
2
Answer:
x2−5x+6x+1
x2−5x+6>0
x2−3x−2x+6>0
x(x−3)−2(x−3)>0
(x−3)(x−2)>0
x∈[−∞,2)∪(3,∞)
Domain of
x+1∈R
∴ Domain of x2−5x+6x+1 is R−(2,3)
= R−(2,3)
HOPE ITS HELPFUL
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